EINSTEIN’S EQUATION IN THE VECTOR MODEL FOR GRAVITATIONAL FIELD Vo Van On University of Natural Sciences, VNU-HCM Manuscript received on August 05 h , 2006; Manuscript received on July
Trang 1EINSTEIN’S EQUATION IN THE VECTOR MODEL
FOR GRAVITATIONAL FIELD
Vo Van On
University of Natural Sciences, VNU-HCM
(Manuscript received on August 05 h , 2006; Manuscript received on July 11 th , 2007)
ABSTRACT: In this paper, based on the vector model for gravitational field we deduce a
equation to determinate the metric of space- time This equation is similar to Einstein‘s equation The metric of space – time outside a static spherical symmetric body is also determined It gives a small supplementation to the Schwarzschild metric in the General Theory
of Relativity but no singular sphere exists
1 INTRODUCTION
From the assumption of the Lorentz invariance of gravitational mass, we used the vector model to describe gravitational field in the non- relativistic case and the relativistic one [1] In these descriptions, space- time is flat yet because we did not consider to the influence of gravitational field upon the metric of space- time yet From the previous paper [2], we have known that the field of inertial forces is just the field of gravitational force and moreover space- time is curvature with the present of inertial forces [3] Therefore space – time also becomes the curvature one with the present of gravitational field
In this paper we shall deduce a equation to describe the relation between gravitational field,
a vector field, with the metric of space- time This equation is similar to Einstein‘s equation We
say it as Einstein‘s equation in the vector model for gravitational field
This equation is deduced from a Lagrangian which is similar to the Lagrangians in the vector – tensor models for gravitational field [4,5,6,7] Nevertheless in those models the vector field takes only a supplemental role beside the gravitational field which is a tensor field The tensor field is just the metric tensor of space- time Those authors wanted to homogenize the vector field with the electromagnetic field In this model the gravitational field is the vector field and its resource is gravitational mass of bodies This vector field and the energy- momentum tensor
of gravitational matter determine the metric of space – time The second part is a Einstein’s essential idea and it is required so that this model has the classical limit
In this paper we also deduce a solution of this equation for a static spherical symmetric body The obtained metric is different to the Schwarzschild metric with a small supplementation of high degree and no singular sphere exists
2 LAGRANGIAN AND FIELD EQUATION
We choose the following action
S =S E +S Mg +S g (1)
with S H−E =∫ −g(R+Λ)d4x is the classical Hilbert –Einstein action
S Mg is the gravitational matter action
Trang 21 4
16
π
= ∫ − is the gravitational action
Where E gμν is tensor of strength of gravitational field
Variation of the action (1) with respect to the metric tensor leads to the following modified Einstein’s equation
G
c
Note that
- Variation of the Hilbert – Einstein action leads to the left- hand side of equation (2) as in the General Theory of Relativity
- Variation of the gravitational matter action S Mg leads to the energy- momentum tensor of
the gravitational matter . 2 Mg
Mg
S T
g g
δ δ
≡ −
−
- Variation of the gravitational action Sg leads to the energy- momentum tensor of gravitational field .
g
S T
g g
δ δ ω
≡ −
−
Let us discuss particularly to two tensors in the right – hand side of equation (2)
We recall that the original Einstein’s equation is
μν μν μν π μν
T c
G g
R g
2
− (3)
Where Tμν is the energy- momentum tensor of the matter For example, for a fluid matter of non- interacting particles with the proper inertial mass density ρ0(x), with a field of 4- velocity uμ(x)and a field of pressure p (x) , the energy- momentum tensor of the matter is [8,9]
2 ( )
μν ρ c u u p u u g
T = + − (4)
If we say ρ 0 as the gravitational mass density of this fluid matter, the energy- momentum tensor of the gravitational matter is
T Mgμν =ρg0c u u2 μ ν +p u u( μ ν −gμν) (5)
For the fluid matter of electrically charged particles with the gravitational mass ρ 0 , a field
of 4- velocity uμ(x), and a the electrical charge densityσ0(x), the energy – momentum tensor
of the gravitational matter is
Trang 30 2 1 1
Tμν ρ c u uμ ν F Fαμ αν g F Fμν αβ αβ
π
= + − + (6)
The word “g” in the second term group indicates that we choose the density of gravitational
mass which is equivalent to the energy density of the electromagnetic field Where Fαβ is the
electromagnetic field tensor
Note that because of the close equality between the inertial mass and the gravitational
mass, the tensor Tμνis closely equivalent to the tensor T Mg.μν The only distinct character is that
the inertial mass depends on inertial frame of reference while the gravitational mass does not
depend one However the value of ρ0in the equation (4) is just the proper density of inertial
mass, therefore it also does not depend on inertial frame of reference Thus, the modified
Einstein’s equation (2) is principally different with the original Einstein’s equation (3) in the
present of the gravitational energy- momentum tensor in the right-hand side
From the above gravitational action, the gravitational energy-momentum tensor is
g
S
g g
δ
ω
Where Eg.αβ is the tensor of strength of gravitational field [1] The expression of (7) is
obtained in the same way with the energy- momentum tensor of electromagnetic field
Let us now consider the equation (2) for the space- time outside a body with the
gravitational mass Mg (this case is similar to the case of the original Einstein’s equation for the
empty space) However in this case, the space is not empty although it is outside the field
resource, the gravitational field exists everywhere We always have the present of the
gravitational energy-momentum tensor in the right-hand side of the equation (2) When we reject
the cosmological constant Λ, the equation (2) leads to the following form
.
1
Rμν − g Rμν = ω T μν (8)
or :
Rμν g Rμν ω E Eαμ αν g E Eμν αβ αβ
π
3.THE EQUATIONS OF GRAVITATIONAL FIELD IN CURVATURE SPACE- TIME
We have known the equations of gravitational field in flat space- time [1]
∂kEg mn. + ∂mEg nk. + ∂nEg km. = 0 (10)
and
k
g
ik g
∂ (11)
Trang 4The metric tensor is flat in these equations
When the gravitational field exists, because of its influence to the metric tensor of space- time, we shall replace the ordinary derivative by the covariant derivative The above equations become
Eg m n k ; + Eg n k m ; + Eg km n ; = 0 (12)
i g D g ik J g k
1
(13)
4 THE METRIC TENSOR OF SPACE-TIME OUTSIDE A STATIC SPHERICAL SYMMETRICAL BODY
We resolve the equations (9), (12) ,(13) outside the resource to find the metric tensor of space- time Thus we have the following equations
Rμν g Rμν ω EαμE αν g E Eμν αβ αβ
π
Eg m n k. ; + Eg n k m. ; + Eg km n. ; = 0 (15)
∂ ( − ik ) = 0
g
i g E (16)
Because the resource is static spherical symmetrical body, we also have the metric tensor in the Schwarzschild form as follows [8 ]
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
−
−
−
=
θ
λ ν
μα
2 2
2
sin
r r e
e
g (17)
and
⎟⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜⎜
⎜
⎜
⎜
⎜
⎝
⎛
−
−
−
−
−
θ
λ ν
μα
2 2
2
sin
1
r r
e e
g (18)
The left-hand side of (14) is the Einstein ‘s tensor , it has only the non-zero components as follows [ 8,9,10]
e r r r e R g
2
− − (19)
Trang 5ν λ
e r r r R g
2
− (20)
2 2 ) ( 4 4
[ 2
2 2 2
22
22− g R=e− λ r ν′λ′−r ν′ −r ν′′− r ν′−λ′
22 22
33
2
1 ( 2
1
R g R
R g
R − = − (22)
Rμν =0, gμν =0 with μ ≠ ν
The tensor of strength of gravitational field E g.μν when it was corrected the metric tensor needs corresponding to a static spherical symmetrical gravitational fieldEg(r ) From the form
of E g.μν in flat space- time [1]
.
g
E
μν
(23)
For static spherical symmetrical gravitational field, the magneto–gravitational components 0
=
g
Hr
We consider only in the X- direction, therefore the components E gy,E gz =0 We find
a solution of E g.μν in the following form
.
1 ( )
c
μν
−
=
(24)
Note that because Eg.μν is a function of only r, it satisfies the equation (15) regardless of function E g(r).The function E g(r) is found at the same time with μ and và ν from the equations (14) and (16) Raising indices in (24) with gαβ in (18) , we obtain
( )
1 0 0 0
c
μα − + ν λ
=
(25)
and
Trang 6( ) / 2 2
1 0 0 0 1
sin
c
μα − + ν λ θ
(26)
Substituting (26) into (16), we obtain an only nontrivial equation
[ − ( ν + λ / 2 2 sinθ ]′ =0
g
E r
e (27)
We obtain a solution of (27)
e− ( ν + λ / 2r2E gsinθ =constant
or
( /2 2tan
r
t cons e
E g = ν+λ (28)
We require that space- time is Euclidian one at infinity, it leads to that both νandλ → 0 when r→∞ , therefore the solution (28) has the normal classical form when r is larger, i.e 2
r
GM
Therefore constant≡−GM g (29)
To solve the equation (14), we have to calculate the energy- momentum tensor in the right-hand side of it We use (28) to rewrite the tensor of strength of gravitational field in three forms
as follows
. ( )/ 2 2
0 1 0 0
1 0 0 0 1
0 0 0 0
0 0 0 0
g g
GM
ν λ
−
(30)
( )/ 2 2
0 1 0 0
1 0 0 0 1
0 0 0 0
0 0 0 0
g g
GM
μα − + ν λ
(31)
Trang 7
( )/ 2 ( )/ 2
1
g g
e
E
ν λ
λ ν α
μ
−
−
(32)
we obtain the following result
. 1 . . 1 .
kl
T μα E μβEβα g E Eμα
π
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
−
=
θ π
λ ν
2 2
2 4
2
2 2
sin 0
0 0
0 0
0
0 0
0
0 0
0
8
r r e e r c
M
(33)
From the equations (14),(19),(20),(21),(22) and(33), we have the following equations
ν λ ν ν
π ω
λ
e r c
M G e
r r r
2 2 2
1 )
1
− (34)
λ λ
π ω
ν
e r c
M G e
r r r
g
4 2
2 2 2
1
−
′
− (35)
4 2
2 2 2
2 2 2
8 )]
( 2 2 ) ( 4 4
r c
M G r
r r
r
π ω λ ν ν ν
λ
ν
λ ′ ′− ′ − ′′− ′− ′ =
Multiplying two members of (34) with e− ( ν − λ )then add it with (35), we obtain
ν′+λ′=0⇒ν +λ =cons tan t (37) Because both ν andλ lead to zero at infinity, the constant in (37) has to be zero
Therefore, we have ν =−λ (38)
Using (37), we rewrite (36) as follows
2 2 2
2 2 2 2
8 )]
( 2 2 ) ( 4 ) (
4
r c
M G r
r r
r
π ω ν ν ν
ν ν
ν − ′ − ′ − ′′− ′+ ′ =
or
2 2 2
4 ]
2 )
[(
r c
M G r
π ω ν
ν ν
ν ′ + ′′ + ′ = − (39)
2 2 2
4
2 ] )
[(
r c
M G e
r
π ω ν
ν
ν ′ + ′′ + ′ = − (40)
We rewrite (40) in the following form
Trang 82 4
2 2 4
2 )
(
r c
M G e
r
π ω ν
ν ′ ′ + ′ = − (41) Puty = eνν′ , (41) becomes
2 2 4
2
r c
M G y
r
π ω
−
= +
′ (42) The differential equation (42) has the standard form as follows
y′+ p(r)y=q(r) (43)
The solution y (r)is as follows [11]
2 )
( )
We have
= (∫ ( ) ( ) + )
) (
1 )
r r
) ]
4 ( [
4 2
2 2
r c
M G r
−
]
4 [
1
2
2 2
r c
M G r
g +
=
π ω
2 3 2
2 2
A r c
M
+
=
π
ω (45) Where A is the integral constant
From the above definition of y = eνν′, we have
eνν′=(eν)′= 2 3 2
2 2
A r c
M
+
π ω
or
dr
r
A r c
M G
4
2 2 +
=∫ ω π ν
B
r
A r c
M
+
−
−
2 2
8π
ω (46) Where B is a new integral constant
We shall determine the constants A,B from the non-relativistic limit We know that the Lagrangian describing the motion of a particle in gravitational field with the potential ϕghas the form [10]
Trang 9L=−mc + mv −mϕg
2
2 2
The corresponding action is
=∫ =− ∫ − + dt=−mc∫ds
c c
v c mc Ldt
2 (
we have dt
c c
v c
2 (
+
−
=
2 2
2
2 2
4 2
4
c
v v
c c
v c
g
ϕ
ϕ
− +
− + +
= =(c2 +2ϕg)dt2−v2dt2+
(1 2 ) 2 2
2
c
(47) Where we reject the terms which lead to zero when c approaches to infinity
Comparing (47) with the our line element (we reject the terms in the coefficient of dr2)
ds2 = eνc2dt2 − dr2 (48)
we get
− + ≡2 2 +1
c
B r
≡−2 2 +1
r c
GM g
(49) From (49) we have
2 2
c
GM
A= g (50)
B = 1
The constant ω does not obtain in the non relativistic limit because it is in high accurate
terms, we shall determine it later
Thus, we get the following line element
8 2
1 ( ) 8 2
1
2 2
2 2
2 2
2 2
2 2
2 2
π
ω π
r c
M G r c
GM dt
r c
M G r c
GM c
(51)
we put 2
ω π
ω = ′
and rewrite the line element (51)
) sin (
) 2
1 ( ) 2
1
2 4
2 2 2
2 2 4
2 2 2
2
r c
M G r
c
GM dt
r c
M G r
c
GM
c
ds = − g − ′ g − − g − ′ g − − + (52)
Trang 10We determine the parameter ω′ from the experiments in the Solar system We use the
Robertson – Eddington expansion [9] for the metric tensor in the following form
) sin (
) 2
1 ( .) )
( 2 2
1
2
2 2
4
2 2 2
2
r c
GM dt
r c
M G r
c
GM
c
ds = − g+ − g+ − + g+ − + (53)
When comparing (52) with (53), we have
α =γ =1 (54)
and ω′=2(1−β) (55)
The predictions of the Einstein field equations can be neatly summarized as
α =β =γ =1 (56)
From the experimental data in the Solar system, people [43] obtained
1.00 0.01
3
2
⎠
⎞
⎜
⎝
⎛ −β + γ
(57) With γ = 1 in this model, we have
ω ′ = 2 ( 1 − β ) = 0 00 ± 0 06 (58)
Thus ω′ ≤0.06 hence ω ≤ 0 48 π c−2 (59)
The line element (52) gives a very small supplementation to the Schwarzschild line element
It is interesting to note that the function 4 2
2 2 2
2 1
r c
M G r
c
GM
eν = − g − ω ′ g takes the form shown
in Fig.1
Fig.1 the graphic of function eν
In particular, we see that no singular sphere exists in the line element (52), unlike the case
of the ordinary Schwarzschild line element which possesses a singular sphere at 2 2
c
GM
r= g
ν
e
r 1
Trang 115 CONCLUSION
In conclusion, based on the vector model for gravitational field we deduce a modified Einstein’s equation This equation gives a small supplementation to the results of General Theory of Relativity and in particular no singular sphere exists Some different effects of GTR will be investigated later
PHƯƠNG TRÌNH EINSTEIN TRONG MÔ HÌNH VECTOR CHO TRƯỜNG
HẤP DẪN
Võ Văn Ớn
Trường Đại học Khoa học Tự nhiên, ĐGQH-HCM
TÓM TẮT: Trong bài báo này, dựa trên mô hình véctơ cho trường hấp dẫn chúng tôi rút
ra một phương trình để xác định mêtríc của không – thời gian Phương trình này là tương tự với phương trình Einstein mêtríc của không – thời gian bên ngoài một vật đối xứng cầu, dừng cũng được xác định Nó cho một bổ chính nhỏ vào phần tử đường Schwarzchild của Thuyết Tương Đối Tổng Quát nhưng không có cầu kì dị trong nghiệm này